Lời giải:
a. ĐKXĐ: $a\neq \pm 2$

\(M=\frac{(2+a)^2}{(2-a)(2+a)}+\frac{4a^2}{(2-a)(2+a)}-\frac{(2-a)^2}{(2+a)(2-a)}\)

\(=\frac{(2+a)^2+4a^2-(2-a)^2}{(2-a)(2+a)}=\frac{4a(a+2)}{(2-a)(2+a)}=\frac{4a}{2-a}\)

b.

$|a+1|=3\Rightarrow a+1=\pm 3\Rightarrow a=-2$ hoặc $a=-4$

Vì $a\neq \pm 2$ nên $a=-4$

Khi đó: $M=\frac{4a}{2-a}=\frac{4(-4)}{2-(-4)}=\frac{-8}{3}$

c.

Trước tiên cần tìm $a$ để $M$ nguyên đã.

$M=\frac{4a}{2-a}=\frac{8-4(2-a)}{2-a}=\frac{8}{2-a}-4$ nguyên khi $\frac{8}{2-a}$ nguyên 

$\Rightarrow 2-a\in\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$

$\Rightarrow a\in\left\{1; 3; 0; 4; -2; 6; 10; -6\right\}$.

Thử lại thấy $a\in\left\{1; 3; 0; 4\right\}$ thỏa mãn $M$ là số nguyên chia hết cho $4$

a: ĐKXĐ: a<>3; a<>-3; a<>-1

b: \(P=\dfrac{2a^2-3a+3a+9-2a^2-3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}\)

\(=\dfrac{6}{\left(a+3\right)\left(a+1\right)}\)

c: |a|=2

=>a=2 hoặc a=-2

Khi a=-2 thì \(P=\dfrac{6}{\left(-2+3\right)\left(-2+1\right)}=-6\)

Khi a=2 thì \(P=\dfrac{6}{\left(2+3\right)\left(2+1\right)}=\dfrac{6}{5\cdot3}=\dfrac{2}{5}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}-2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)

\(=\dfrac{2\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{2a}\)

\(=\sqrt{a}+2\)

b: A-2<0

=>\(\sqrt{a}+2-2< 0\)

=>\(\sqrt{a}< 0\)

=>\(a\in\varnothing\)

c: Bạn ghi đầy đủ đề đi bạn

a) ĐKXĐ: \(x\ne3;x\ne\pm2\)

\(C=\frac{2a-a^2}{a+3}\cdot\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)

\(C=\frac{-a^2+2a}{a+3}\cdot\left(-\frac{4a}{a-2}\right)\)

\(C=-\frac{2a-a^2}{a+3}\cdot\frac{4a}{a-2}\)

\(C=-\frac{\left(2a-a^2\right)\cdot4a}{\left(a+3\right)\left(a-2\right)}\)

\(C=\frac{4a^2}{a+3}\)

b) \(C=\frac{4.4^2}{4+3}=\frac{46}{7}\)

c) \(\frac{4a^2}{a+3}=1\)

<=> 4a² = a + 3

<=> 4a² - a - 3 = 0

<=> 4a² - 3a - 4a - 3 = 0

<=> a(4a + 3) - (4a + 3) = 0

<=> (4a + 3)(a - 1) = 0

<=> 4a + 3 = 0 hoặc a - 1 = 0

<=> a = -3/4 hoặc a = 1

sửa đáp án câu b thành \(\frac{64}{7}\) nhé

C = 4a^2/a + 3 để C nguyên

=> 4a^2 chia hết cho a + 3

=> 4a^2 - 36 + 36 chia hết cho a + 3

=> (2a - 6)(2a + 6) + 36 chia hết cho a + 3

=> 2(a + 3)(2a - 6) + 36 chia hết cho a + 3

=> 36 chia hết cho a + 3

=> a  + 3 thuộc {-1; 1; -2; 2; -3; 3; -4; 4; - 6; 6; - 9; 9; - 12; 12; -36; 36}

=> a thuộc {-4; -2; -5; -1; -6; 0; -7; 1; -9; 3; -12; 6; -15; 9; -39; 33} mà x khác  - 2

=> a thuộc ...

a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)

\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)

2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)

a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài